text:
In which direction will an aircraft drift after taking off in a crosswind?
Assuming the pilot applies the appropriate crosswind flight controls (see below) during takeoff and rotation, and releases those flight controls after the aircraft leaves the ground, in what direction will the aircraft's ground track diverge from the extended runway centerline as the aircraft accelerates from takeoff velocity?
On a recent practical test, I had a DPE insist to me that the aircraft, as soon as it breaks ground, will immediately begin drifting downwind. We then had an extensive discussion, in the oral portion of my examination, as I am of the (I hesitate to use the word "opinion" because this is basic physics of which I have absolutely no doubt), but let me say I attempted to explain to the DPE that in the scenario described, the aircraft would immediately weathervane into the crosswind, to eliminate the sideslip it was in, and thereafter, as it accelerated, its ground track would deviate upwind of the extended runway centerline, not downwind.
I am posting this because I am curious as to the extent of this misconception among the general aviation community. If DPE, certified by the FAA to administer Practical Tests, has this incorrect idea I wonder how many others do not understand the basic concept that wind is not pushing on aircraft in flight, it is but a mathematical concept representing a translation of values from one frame of reference (the surface of the earth) to another (moving) frame of reference, (the airmass around the aircraft). Also, I thought that a discussion, here in this forum would be useful and educational for all.
NOTE. By proper crosswind takeoff controls, I mean of course, downwind rudder (opposite the crosswind), to keep the aircraft tracking down the centerline, and upwind aileron (into the wind), to prevent the crosswind from lifting the upwind wing ahead of the downwind wing at rotation.  These cross-controls effectively create a slip at rotation, and allow the aircraft to rotate and liftoff with the fuselage aligned with runway centerline even though, (within the moving airmass) it is flying sideways (the fuselage is misaligned with the relative wind). Of course, as soon as the cross controls are released, the fuselage will rotate (weathervane) into the crosswind and the sideslip angle will be eliminated.
Because so many readers do not understand the basic physics underlying the concept of wind, (which is that in the airmass, there is no wind) I have decided to present this thought experiment.

 Imagine you are flying North towards an airfield with a North-South Runway. You are on extended runway centerline, but there is a strong crosswind from the East. You want to track along the runway centerline directly towards the runway, so you establish the appropriate crab angle by turning the aircraft into the wind.
At point A, then, the aircraft ground track is now Due North, perfectly aligned with the runway centerline. The aircraft heading, of course, is slightly to the right of North, by whatever crab angle is required to counteract the movement of the air mass (the "Wind").
Now, at Point B, you decide you might want to land. To minimize the sideloads on the landing gear (or for whatever other reason you may imagine), you want to align the fuselage of the aircraft with the runway, WITHOUT changing your ground track. So you add in the appropriate amount of left rudder to bring the nose to the left so as to align the aircraft fuselage with the runway heading (North). Now, at point C,  the aircraft heading, (AND THE AIRCRAFT GROUND TRACK), are aligned with the runway. They are both True North. Because the aircraft is in a left sideslip (or left forward slip), however you want to describe it), adverse yaw due to the dihedral effect will cause the aircraft to slowly turn, (change its heading) very slightly, to the left. To stop this you add a small amount of right bank, which will require some right aileron to maintain. The aircraft is now in a cross-controlled sideslip, but it is stably tracking across the ground due North, with its fuselage and heading aligned due north with the runway.
Now, and here's the question that determines if you understand this or not. Halfway down the runway you neutralize the controls. You are simply returning the flight controls to the same place they were at Point A though Point B. What will happen?
The answer, is that after neutralizing the controls, the aircraft, in EVERY WAY YOU CAN POSSIBLE EXAMINE, is in EXACTLY the same aerodynamic conditions it was in from Points A to Point B.
So, if you still maintain that the aircraft will immediately take up a ground track to the left as is depicted in the diagram along path A, and not simply weathervane into the relative wind coming from the right of its nose, and continue to track straight ahead, along path B, you are wrong, but more than that, you have to be able to explain what is different about the situation after neutralizing the controls at Point D, than between Point A and Point B.
The only thing that can possibly explain someone maintaining this belief is the false concept that the wind is Pushing on the side of the aircraft, and that this (fictitious and non-existent) force is what causes wind drift.
Nothing could be further from the reality. There is no force. Bodies only change their velocity when a force is applied. $F=ma$ is true, always was true, and always will be true. Since "wind" is fictitious, it cannot exert a force on anything free to move in the airmass. Since there can be no force from wind (it is only an abstraction representing the difference between measurements in two frames of reference moving with respect to one another, there can be no change of velocity due to a wind. The law of inertia says that bodies will always continue to move at a constant velocity unless a FORCE is applied. Without a force, the aircraft will continue to maintain a constant inertial velocity.

Are you actually asking a question, or do you just want people to tell you you're right?

Gutsy move arguing with a DPE, and I'm curious if you've validated your theory empirically. Anyway, to make this question clearer, it might help to spell out exactly what you believe the "appropriate crosswind controls" are. It's not clear - to me! - what what technique you're describing, and "weathervaning" is usually used to describe something that happens on the ground, not in the air.

@Dan, I guess I am doing exactly what the question says, polling to see what others think. And I think the topic is of interest, and that a discussion would be useful to all.

@Michael, when taking off in a crosswind, with proper takeoff crosswind flight controls applied, immediately after liftoff, the aircraft is in a sideslip, exactly as you would be in just prior to touchdown on a landing.

Annotated the question to describe the proper crosswind takeoff controls

@ymb1,  Noted, agreed, and Done. Thank you.

The DPE is right, and this is incredibly easy to verify. On your next flight take off in a cross-wind and then center the controls. The aircraft weathervanes into the wind, yes, but look at the runway and notice that you are physically ground-tracking in the downwind direction. The only way to track upwind is to turn more into the wind, your aircraft will drift with the air mass it is flying in. This is basic piloting, and exactly why you need to take wind into account when computing headings to a destination.

@Ron, I have done this many times, and I have seen this effect many times. When the aircraft weathervanes into the wind, it HAS turned (fuselage heading) into the wind. That is exactly what weathervaning means. When you think about turning (when anyone talks about "Turning"), it is critical to not conflate turning (rotating or yawing only the fuselage and the mag heading) from the turning that changes the aircraft velocity vector in inertial space (across the ground).

@CharlesBretana I'm not sure why you are arguing this, I'm a pilot and have been for a long time. The DPE is a pilot and has been for a long time, the people who have commented here have been flying a long time. My (and others) experience differs from yours. I'm sure your instructor would agree, those offering different points are not wrong. I understand your conviction, but it isn't based in reality. I think you are confusing your "air track" with your "ground track".

@Ron, I have been a pilot for a long time too. I am an aeronautical engineer, with a masters degree in engineering. and I understand this topic thouroughly. I taught it in the USAF as part of tactical fighter ground attack academic training, and experienced the same issues there, as those without a fundamental understanding of what wind really is, approached this topic with the exact same misconceptions that wind pushes on the side of the aircraft, rather than simply representing a  frame translation.

In the airmass, there is no crosswind, so it cannot exert a force on the side of aircraft to change it's ground track. There is only the Relative WInd, and all it can do is change the alignment of the aircraft fuselage. The wings, when the aircraft is banked, is the way to change aircraft ground track. Crosswind cannot do that. (rudder or sideslip has a very small effect, but it is insignificant compared to the effect of the wings in a bank) I believe it is you who may be confusing "air track" with "ground track" .

I won't compare endowments here but suffice it to say that I'm also an engineer who works on guidance systems (among other things)... This isn't going anywhere so I'm going to pass on further discussions on the topic, good luck.

I tried to have this discussion analytically, without bringing in endowments, (as you describe them), but you brought in your experience. My apologies for not simply agreeing with you because "*(you have been a)... pilot and have been for a long time. ... *"  I have attempted to make numerous substantive analytical points about this, but few comments have addressed or refuted them. It's sad, but it seems that this idea, at least among those who responded, may be too deeply entrenched to allow a rational examination.

Let us continue this discussion in chat.

In a theoretical scenario of constant wind, you are correct in that if you trim for a steady crosswind immediately after takeoff and then accelerate, your track deviates upwind, as the “sideways component” of your still air velocity vector after accelerating exceeds the cross wind. In most real scenarios, however, I’d expect wind to vary with altitude in both direction and velocity, so I don’t believe there would be many opportunities to actually observe the upwind travel scenario, and the aircraft will most commonly track downwind when heading remains constant.

I think this question may be improved.  Re " Because the aircraft is in a left sideslip (or left forward slip), however you want to describe it), adverse yaw due to the dihedral effect will cause the aircraft to slowly turn, (change its heading) very slightly, to the left." -- I don't think that this is really "adverse yaw".  Wouldn't it be more clear to say that "the dihedral effect generates a left roll torque which would lead to a left bank, causing a left turn.  Even with the wings level, the airflow impacting the right side of the fuselage will generate an aerodynamic sideforce....

(Ctd) ...Even with the wings level, the airflow impacting the right side of the fuselage will generate an aerodynamic sideforce that will tend to cause a left turn.  To stop this you add a small amount of right bank, which will require some right aileron to maintain, assuming that the aircraft has some amount of coupling between slip and roll."

The DPE is correct, the airplane will drift in the direction of the wind.
If you are piloting a boat across a river, and point the bow at the opposite side of the river with the rudder amidships then the boat will float in the direction of the current and stay where you pointed it. The boat is not going to turn upstream by itself because the water exerts equal force against the bow and stern.
The same physics is at work in an airplane, air exerts an equal force on the front and rear of the airplane, it's not going to somehow blow harder on the tail.

"it's not going to somehow blow harder on the tail" Well, technically, I suppose it can, but hopefully not for long enough to matter for the purpose of what OP is describing. Think downdrafts, windshear, and similar weather phenomena which can cause very distinct shifts in wind passing through various boundary layers (both vertical and horizontal).

Building turbulence can yaw you around for sure, as I have experienced plenty of times @MichaelKjörling, but as you say never for very long. I may refine this answer a bit, I wanted to keep it simple.

If the aircraft is in a sideslip, with no rudder deflection to keep it there, then that's EXACTLY what it does. The wind blows on the rudder from the side, creating more force on the back of the aircraft then on the front. (This is why aircraft designers put the rudder at the BACK).

Wrong, (about the boat), the DPE also tried the boat analogy. Boats have more or less equal cross surfaces exposed to the "sideslip" forces created by a cross-current, and a great deal of rotational inertia that resists weather-vaning. AIrcraft (the RUDDER) have a great deal of lateral asymmetry that creates a weathervaning torque, and much lower rotational inertia. Think what happens when you are cross-controlled in a forward slip to a landing, and suddenly neutralize the controls). Better yet, just look behind you the next time you take off in a crosswind.

@CharlesBretana, your position contravenes physics and aerodynamics as taught and my own 15 years of piloting with hundreds of crosswind departures, however I'd be happy to accept it's wrong if you can put together a case for it. Or perhaps you haven't expressed your question in a way the community understands, either way I'd say you have some work to do to prove your point.

@GdD, I apologize for seeming adamant, but please check your physics. The topic is rotational moments. If you apply a force against an object free to rotate about it's center of gravity, at any point other than its CG, it will cause the object to rotate about the CG. The reason the rudder is placed at the back of aircraft is to create exactly such a force. It keeps the aircraft fuselage laterally aligned with the relative wind. This is exactly the situation at liftoff in a crosswind.

(con't) At liftoff, the fuselage is aligned with the runway centerline, yet the relative wind will be hitting the aircraft from the upwind side. This creates a sideslip angle, which creates the lateral yaw moment, that will align the fuselage with the relative wind at that moment in time. Now the aircraft is pointed upwind. If it maintained a constant velocity, it would track straight along the runway centerline, but as it accelerates, its ground track will diverge upwind of the extended runway centerline.

I know my physics @CharlesBretana, and I know why the rudder is at the back. I think what you are describing is actually happening due to your own control inputs and/or gyroscopic/slipstream effects creating unbalanced flight, which is potentially hazardous in slow flight. Transitioning between crossed controls and balanced flight is a skill that comes with practice, concentrate on keeping the ball in the middle.

@GdD, What do you think :"Keeping the ball in the middle" means? If you rotate and lift off with the appropriate cross-controls for a significant (for any) crosswind, are you saying that you think the ball will be centered, and that to keep it centered you need to maintain those cross-controls? The control inputs I am describing are simply releasing the cross-controls. That WILL center the ball.

@CharlesBretana, releasing the controls will not center the ball, you need rudder to do that! Please, please be willing to learn from others and accept you may not understand this fully. In any case extended discussion is discouraged, feel free to open a chat with the community.

@GdD, I am not being difficult, I am responding to your comments carefully and analytically. I would hope that you will at least grant me the boon of thinking about this before responding from preconceived attitudes. Think, If you establish a significant sideslip (Full rudder in one direction and sufficient opposite aileron to maintain a constant heading), and suddenly neutralize both rudder and aileron, What happens to the ball?

@GdD -- re "releasing the controls will not center the ball" -- why not?  Why is the ball off center, unless the rudder is being applied to expose the side of the fuselage to the airflow and generate an aerodynamic sideforce, despite the fact that the vertical fin is trying to "weathervane" the nose into alignment with the airflow and eliminate this sideforce?  Sure we could get into P-factor etc, but those are second-order effects compare to the main topic at hand.

You drift downwind unless you turn into it.  In a normal crosswind takeoff, say with the wind from the left, at rotation, you have left aileron to keep the left wing from lifting and right rudder to maintain runway heading during the rotation.  You are taking off in a sideslip.  As soon as you are clear, you center the controls to take out the sideslip while still on the runway heading and you now are in ball-centered coordinated flight, on runway heading through an air mass moving from left to right.
The airplane won't weathervane into the surface wind because it's no longer anchored to the ground; it weathervanes into its own relative wind as it normally does.  You are now on runway heading in coordinated flight in an air mass moving from left to right.  The airplane's relative wind is straight ahead if the ball is centered.  Where is the airplane's track?  To the right of the heading.  You will see it as a drift downwind of the runway centre line as you climb out.  There may be a very brief weathervaning tendency in the instant following weight off wheels as inertia resists the lateral drift, but that is quickly overcome.
The proper technique on a VFR takeoff is to perform a coordinated turn into the crosswind once airborne to establish a crab angle that will allow you to maintain the runway track.
The only time you don't do this crab correction is on an IFR takeoff where you are given a departure heading.  You maintain that heading and let ATC worry about any drift.

Comments are not for extended discussion; this conversation has been moved to chat.

@JohnK -- re "As soon as you are clear, you center the controls to take out the sideslip while still on the runway heading and you now are in ball-centered coordinated flight, on runway heading" -- a sideslip involves being yawed to point in a different direction than the actual direction of travel through the airmass.  If you are intitially on the runway heading but in a sideslip, and then you take out the sideslip and align the nose with the actual direction of travel through the airmass, you won't be on the runway heading anymore.  You'll be pointing upwind.

And as you accelerate, the ground track will veer in the upwind direction.

"There may be a very brief weathervaning tendency in the instant following weight off wheels as inertia resists the lateral drift, but that is quickly overcome".  What  I mean is the crab that results from the initial weathervaning is not sufficient to fully correct for the crosswind and you will still drift downwind if you don't make a conscious heading change to get to the full crab angle required.

@quietflyer, you clearly understand frame translation and I thank you you for supporting me on this thread!

Assuming the pilot applies the appropriate crosswind flight controls
  (see below) during takeoff and rotation, and releases those flight
  controls after the aircraft leaves the ground, in what direction will
  the aircraft's ground track diverge from the extended runway
  centerline as the aircraft accelerates from takeoff velocity?
NOTE. By proper crosswind takeoff controls, I mean of course, downwind
  rudder (opposite the crosswind), to keep the aircraft tracking down
  the centerline, and upwind aileron (into the wind), to prevent the
  crosswind from lifting the upwind wing ahead of the downwind wing at
  rotation. These cross-controls effectively create a slip at rotation,
  and allow the aircraft to rotate and liftoff with the fuselage aligned
  with runway centerline even though, (within the moving airmass) it is
  flying sideways (the fuselage is misaligned with the relative wind).

If the pilot uses "crossed controls" to establish a bank angle sufficient to fully counteract the sideforce from the airflow striking the side of the fuselage before lifting off the ground, while keeping the nose aligned with the runway heading, then there will be no tendency to "drift" either upwind or downwind of runway centerline.
Note that the amount of "upwind" aileron you will need to hold depends strongly on the aircraft's yaw-roll (or more properly slip-roll) coupling characteristics-- with a flat wing mounted at mid-level, you might not need to hold any upwind aileron at all to hold the required bank angle.
What happens if you then release all the controls simultaneously?
When you release the rudder, the aircraft will yaw into the relative wind to align itself with the flight path through the airmass.  When you simultaneously release the ailerons, the aircraft at the first instant will still be experiencing some sideslip (because it has not yet fully yawed into the relative wind) which will create a roll torque toward wings-level (if the slip-roll coupling is non-zero, i.e. if you had to hold any aileron input in the slip at all), but it seems likely that the aircraft will end up being yawed to point directly into the relative wind before the bank angle has reached zero.  At this point the aircraft will be doing a normal, banked, nearly-coordinated turn in the upwind direction.  The aircraft will initially fly upwind, but may continue through multiple 360-degree turns, eventually ending up far downwind of the runway centerline.  Naturally, a pilot would normally make control inputs as needed to prevent all this, rather than simply releasing controls immediately after take-off.
However, perhaps what the DPE was getting at is this-- as airspeed increases, to hold a constant ground track, the "crab" angle must be reduced.  (Note that "crabbing" has nothing whatsoever to do with slipping-- in "crabbed" flight, the aircraft is not flying sideways through the air.)  Starting with the aircraft off the runway and established in wings-level fully coordinated (non-slipping) flight, "crabbed" as needed so that the ground track remains aligned with the runway centerline, then as the airspeed increases, the pilot must make a turning correction in the downwind direction (presumably by banking slightly) to prevent the ground track from veering off centerline in the upwind direction.  If the DPE was saying the opposite, he was certainly mistaken.
When in actual practice we see an aircraft drift toward the downwind edge of the runway after lift-off, it is because the pilot was not able to establish a strong enough upwind correction (by banking in the upwind direction, with the downwind wheel off the ground) before liftoff.  As a result of the inadequate bank angle, there was a sideload on the tires up until the moment of liftoff.  Remove the sideload on the tires by lifting off the ground, and the aircraft starts drifting downwind (or more properly turning downwind, due to the sideforce from the airflow impacting the side of the aircraft) until it reaches equilibrium with the airmass.  All of which could have been prevented by using the proper, steeper, bank angle at the moment of liftoff to bring the sideload on the tires to zero, combined with downwind rudder as needed to keep the aircraft heading aligned with the runway centerline.
It may be the case that the technique of establishing an adequate bank angle before takeoff that is sufficient to eliminate all sideways drift at the instant after takeoff is only possible in aircraft with relatively narrow-track landing gear, such as a typical high-wing general aviation airplane.  In an aircraft with a wider-track gear, perhaps the the maximum bank angle that is reasonably attainable by the instant before takeoff is often shallow enough that some downwind "drift" (turning of flight path due to sideforce of airflow against fuselage) is inevitable.  (This could be the topic of another ASE question.)  It isn't clear whether the constraints of the original question ("the pilot applies the appropriate crosswind flight controls") should be interpreted to encompass this situation or not.  Such a case would still meet the description of "downwind rudder (opposite the crosswind), to keep the aircraft tracking down the centerline, and upwind aileron (into the wind), to prevent the crosswind from lifting the upwind wing ahead of the downwind wing at rotation."
Of course, once the aircraft is off the ground, the usual strategy is to change from the cross-controlled slip to coordinated, crabbing flight with the ground track parallel to the runway centerline.  Whether or not this transition should be considered to be a "turn"-- i.e. a change in the direction of the flight path and ground track-- depends on whether or not the ground track has been allowed to "drift" (turn) downwind immediately after liftoff.  Also, as noted earlier, if the aircraft is accelerating, even if no downwind "drift" (turning of the ground track) has occurred, a slight rate of turn in the flight path in the downwind direction required to maintain a linear ground track.
So-- "it's complicated".  If DPE asked me a similar question, I would ask that the constraints at play be clarified in detail before answering.

Your opinion is a theory. DPE is correct. In the real world, you lift off, crab into the wind and will still move downwind of the centerline. You see this over and over, no airplane ends up on the upwind side of the extended centerline.

Of course some do when they overcompensate. I witnessed one which strayed into a glider area this way and collided with a glider, right after take-off.

No, an aircraft with 5 degrees of crab in a wind requiring only 3 degrees of crab, will move UPWIND. You cannot seriously believe that aircraft will move downwind of a fixed ground reference when it is heading, (pointed), well upwind of that ground reference line.  But your statement above indicates you don't even understand the real issue here. It is not whether an aircraft in a "crab" will move upwind or downwind,  It is whether an aircraft will weathervane into an existing relative wind when it breaks ground on liftoff, and by how much will it weathervane.

You might as well state that an aircraft will move "downwind"of its ground track. Statements like this or like yours only indicate a failure to understand the fundamental principals of physics (not a theory) behind what is nothing more than a transformation between two moving frames of reference.

I believe I see where the confusion is so I will attempt an answer to the question.
This Wikipedia article describes the weathervane effect:

Aircraft on the ground have a natural pivoting point on an axis through the main landing gear contact points [disregarding the effects of toe in/toe out of the main gear]. As most of the side area of an aircraft will typically be behind this pivoting point, any crosswind will create a yawing moment tending to turn the nose of the aircraft into the wind.2[3]

The keyword here is that there is a pivoting point which allows rotation.  When the plane has a pivot point it actually acts as a really good weathervane because more of the force from the wind is exerted on the large vertical stabilizer and much less force is exerted on the nose.  See this cool picture of a DC-3 acting as a weather vane at Yukon Transportation Museum:

However when there is no pivoting point (the plane is in the air) things now change.  The side forces on the vertical stabilizer will still be greater than the nose but with no fixed point of rotation the airplane will slip downwind (hence the term slip in aerodynamics) so the ground track would move downwind (hence the DPE is right I am afraid).  The nose would eventually line up with the relative wind (if the forces remain constant) but not before the plane is blown off course.  The rudder is essentially  used to yaw the plane into the relative wind (which with a crosswind would be angled away from your ground track) to prevent the slip from happening and preventing you from being blown off course.  The technique to use the rudder to correct for the crosswind is called crabbing and is essentially a way to stay in coordinated flight when there is a crosswind to your ground track.
In general the rudder is used to align the nose with the relative wind and bank is used to turn the aircraft in relation to the ground track.  When the nose is aligned to the relative wind the plane is in coordinated flight.

in the air, any body has a natural pivot point, about which any & all applied moments will cause the body to rotate. This point is called the center of Gravity. The law of conservation of momentum requires that unless an external force is applied, the total momentum must remain constant. The total momentum is the product of the velocity of the body tines the mass of the body. If the CG was allowed to change it's velocity the total momentum would be changing. Without an applied force, that would violate the momentum conservation law.

When a force is applied all across the surface of a body in free space, (like from a relative wind not aligned with the fuselage), it's change in momentum from that force affects the trajectory of the CG, its attitude, or orientation is affected by the relative distribution of the forces - forces applied in front of the CG will cause a rotational moment away from the direction of the applied force, force applied behind the CG will cause the nose to rotate n the same direction as the applied force.

in the air, the CG IS the "Fixed point of rotation". Your statement "... use the rudder to correct for the crosswind is called crabbing and is essentially a way to stay in coordinated flight when there is a crosswind ..." is wrong. The rudder MIS-aligns the aircraft with the relative wind. It aligns the aircraft with some ground reference line that is not the same as the natural ground track when the air mass is moving laterally across the ground under the ground track. Use of the rudder (except when used to correct for adverse yaw), creates uncoordinated flight, not coordinated flight!

@DLH -- IF an a/c is experiencing a sideways component in the relative wind, then it will tend to pivot to face into the relative wind, even if it is actually flying, as opposed to being impaled on a pole.  Yes, the same sideways component in the relative wind would exert a sideforce that would tend to turn the flight path in the downwind direction, as you suggest.  But the question is, WHY is that sideforce not fully cancelled out by the bank angle-- both BEFORE the pilot releases pressure on the rudder, and AFTER he releases pressure on the rudder and allows the a/c to weathervane?


text:
How can using split throttles help when landing twins in crosswinds?
In General Aviation light twins, when landing in heavy crosswinds, what are some considerations to using asymmetric (or Split) throttles to induce a sideslip, to align the fuselage with  the ground track and the runway centerline instead of rudder?
I was told that this technique allows the aircraft to track down the runway centerlines with fuselage aligned with the runway without as much bank angle into the crosswind as would be required with symmetric thrust using cross-controlled rudder and opposite aileron alone.
I was thinking that the need for bank into the crosswind is due to the  need to compensate for the sideslip force (from the yaw due to downwind rudder) which would otherwise (without the bank) turn the aircraft downwind.
But it seems to me that using asymmetric thrust to induce the same sideslip would produce the same tendency to turn if the wings were not banked into the crosswind to compensate. Why not??

related: this two flights in which differential thrust was used to control the aircraft trajectory.

The use of asymmetric thrust when landing a twin in a crosswind effectively accomplishes the same thing that rudder input does.  It induces a yawing moment about the aircraft’s vertical axis, the same as pushing on the rudder pedals does.  It helps to alleviate excess rudder pressure on the flight controls during the approach, making it a little more comfortable for the pilot to fly the approach. You are still required to apply aileron input into the wind in order to hold the runway centerline.   Using asymmetric thrust during a Crosswind Landing does not alleviate aileron pressure or required bank angle to track the runway centerline.

That's as I thought. The only difference would be that when using asymmetric thrust, the rudder would not be as deflected. The vertical stabilizer (as well as the entire fuselage because of the sideslip angle) would still be producing lateral aerodynamic force in the downwind direction. So there would still be a need to drop the upwind wing to counteract that force to maintain a constant heading.

@CharlesBretana --  Actually there would be an even greater need-- see new answer--

You may mention response time: pedals/rudder react quicker than engines.

In General Aviation light twins, when landing in heavy crosswinds,
  what are some considerations to using asymmetric (or Split) throttles
  to induce a sideslip, to align the fuselage with the ground track and
  the runway centerline instead of rudder?
I was told that this technique allows the aircraft to track down the
  runway centerlines with fuselage aligned with the runway without as
  much bank angle into the crosswind as would be required with symmetric
  thrust using cross-controlled rudder and opposite aileron alone.

The opposite is actually true-- using differential thrust instead of rudder to hold heading in a wing-down final approach and landing actually creates a slight increase in the bank angle required to hold the aircraft on the desired track.
Consider final approach at a fixed airspeed, with a crosswind from the left.  Consider that we are using the wing-down (slipping) method, aiming to hold the fuselage completely parallel to the runway centerline throughout the last part of the final approach.  The angle between the relative wind and the fuselage is constrained to a fixed value.  The component of aerodynamic sideforce toward the right that is created by the airflow impacting the left side of the fuselage is also constrained to a fixed value.  However, if both throttles are set the same, then to hold the heading constant, we need to deflect the rudder to the right to counteract the left "weathervane" yaw torque from the sideways airflow against the vertical fin.  The deflected rudder contributes an aerodynamic sideforce toward the left, which reduces the overall bank angle required.  (To prevent the aircraft from turning, the bank angle must be sufficient to create a horizontal component of lift to the left that is sufficent to overcome the horizontal component of the net aerodynamic sideforce toward the right.)  If we hold heading by increasing the power on the left engine, so that we can relax the right rudder input, the leftwards aerodynamic sideforce from the deflected rudder vanishes, so we'll need to use more left bank, not less.
In theory, we could reduce the required bank angle by applying differential thrust in the other direction, so that more right rudder is required, but this is undoubtedly never done in actual practice.
Clearly, the real purpose of the differential thrust is simply to relieve the pilot of the need to hold downwind rudder, i.e. to deflect the rudder opposite to the bank angle.  How significant is the resulting increase in the required bank angle?  Or in other words, how significant is the reduction in the required bank angle if the pilot uses the rudder rather than differential thrust in the crosswind landing?
Consider the situation of a failed engine.  In an engine-out situation, the pilot is typically advised to bank about 5 degrees--rarely much more than that-- into the good engine to cancel the sideforce from the rudder and allow linear flight with the fuselage streamlined to the airflow.
If the wing-down sideslip for the crosswind landing requires substantially less rudder deflection than would be needed to compensate for a failed engine at that same airspeed-- as is undoubtedly the case-- then it follows that the reduction in bank angle achieved by maintaining the sideslip with the rudder rather than with differential throttle will be substantially less than 5 degrees. In words, not very significant.  It is clear why a pilot might choose to accept this small increase in bank angle, to relieve the need to hold downwind rudder.
The sideforce from the deflected rudder itself is clearly very small compared to the sideforce from the airflow striking the side of the fuselage as the aircraft maintains the slip angle required for the wing-down crosswind landing.
Note also that "loading up" the lowered wing while "unloading" the raised wing (via the differential power setting and resulting difference in propwash velocity) will slightly increase the aileron input needed to equalize the lift vector from each wing and bring the net roll torque to zero and hold the desired bank angle.
The bottom line is that a pilot using differential thrust to reduce or eliminate the required rudder deflection during a wing-down approach to a crosswind landing is accepting a slight increase in the required bank angle, for the sake of going easy on the muscles in his downwind leg.
Related -- Why does a pilot bank up to 5 degrees into the operating engine following failure of the other engine?

Disclaimer: I don't fly twin-engine aircraft.

In your case (wind from left), with aileron input to induce a left-yaw, the purpose in increasing in throttle on the left engine is to increase lift on the left-wing while counteracting the yaw reducing the amount of rudder input needed to maintain alignment with the center-line. More lift -> higher the force provided to counteract the crosswind with the same bank angle. (AMEL-Instrument)

@DavidC.Rankin -- increasing lift on the left wing in particular would seem to increase, not decrease, the left aileron input required to hold the bank angle.  Total lift cannot exceed vector sum of aircraft weight plus whatever horizontal component of lift is required to neutralize aerodynamic sideforce.  If the aircraft has any dihedral, it would seem increasing lift on right wing, not left wing, would be the most effective way to increase sideforce to left without increasing bank angle.

@DavidC.Rankin -- after giving your comment some thought I still don't see how increasing power on the left engine would decrease the required bank angle to deal with the crosswind from the left.  Plus given the whole issue of the sideforce from the rudder as outlined in my answer, it seems counterproductive-- the required bank angle would actually have to be increased, compared to what would be needed if both engines were run at the same throttle setting.

@DavidC.Rankin --  It seems to me that the ONLY thing that can be accomplished by increasing thrust on the upwind engine is a decrease in required rudder deflection.    What am I missing?

You have to understand the increase in lift due to another 100-200 RPM isn't a huge increase. The primary benefit is the increased thrust creating a moment to balance the yaw induced by aileron inputs (which is what you would do with increased rudder input if you didn't have an engine to play with on each wing).

@DavidC.Rankin -- I don't dispute that the differential thrust will reduce the required rudder input.  My point is that this cannot reduce the required bank angle, and in fact, it must slightly increase the required bank angle.

You are obviously thinking correctly. The question is one that really depends on the flight characteristics of the airplane. Recall in flight dynamics, yaw and roll are coupled. The single-engine cross control with the rudder produces significant tail-drag due to rudder deflection. The split-throttles counters that in 2 ways (1) the increased thrust/lift on the down wing and (2) reduced drag by lessening rudder deflection. Now "how many degrees of bank" difference that results in would take a good model or flight testing to confirm, An Apache will differ greatly from a PBaron.

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Are there any situations where having high lift but low lift to drag ratio would be beneficial?
From my understanding, high lift to drag ratio could make an aircraft efficient during cruising conditions. This is when the aircraft is in equilibrium, lift is equal to weight, and thrust is equal to drag, and since there's less drag, less thrust is required. Is this correct and are there situations where having higher lift but higher drag as a consequence would be beneficial (e.g. higher camber has higher lift, but drag increases more leading to a lower lift to drag ratio)?
For example, if lift generated were to be around 10000N with 700N of drag (14.29 l/d) compared to less lift and drag, 8000N of lift with 600N of drag (13.33 l/d), would less drag at the expense of less lift be worth it in the context of cruising conditions (where thrust would need to equal drag to maintain constant velocity)?
Sidenote: I'm mainly thinking about this in terms of cruising conditions, not necessarily during climb or descent, but if you're happy to share any info on that, I would appreciate it too!
EDIT: Just to clarify (thanks for the comment quiet flyer), this is lift and drag generated by the wings.

This question could use some clarification.  There are certainly cases where having a high lift coefficient, but a low L/D ratio, would allow the aircraft to come in on final approach at a low airspeed but a steep approach path.  Clearly that could be good in some situations.  In cruise-- not likely.  Anyway, if you mean lift coefficient rather than just plain lift as a force in pounds or newtons, you should say so.

Are you sure you want to specify absolute lift and not its coefficient? Realize that, once you take off, the only way to increase required lift is to pull g-forces (e.g. in a tight turn). So your question is really, "is there a situation where high g-forces are pulled where you want a lot of drag?". Is that your intended question?

Alright now I'm actually quite confused about the difference between lift and lift coefficient, but I'm not talking about g-forces, just generated lift by the wings.

Regarding the example you just added: In cruise conditions lift must equal weight and is therefore fixed.

That's pretty much the whole point of landing flaps. Lots of drag to slow you down but high lift to keep your descent rate under control.

I was waiting for the question to be clarified before answering, but it seems that the confusion between forces and their coefficients is not likely to be straightened out any time soon, plus there are now answers posted that would be invalidated by substantial changes, so here goes--

@SharkyPanda, if you are talking about cruise, do you realise that lower absolute lift implies lower weight of the aircraft? It's an absolutely valid thing to ask, but you need to clarify that the conditions are not the same. If, on the other hand, you have the same conditions (i.e. weight) in mind, then, as others have noted, you can't arbitrarily change [absolute] lift. Depending on your choice, one or the other of the already posted answers will be correct.

Landing phase would benefit from high lift but low lift-to-drag ratio. At most phase of flight you need about the same amount of lift to keep the plane in the air. However during landing you need to slow down to landing speed. Hence you lower lift-to-drag ratio by keep the same amount of lift but increase amount of drag. This is usually accomplish by selection of large amount of flaps.

In cruise: No. Less drag means less thrust, which is always beneficial for the practical operation of an airplane.
There is only one condition except for approach and landing where high drag helps, an that is also not during cruise: In aerobatic airplanes during vertical maneuvers.
If, for example, the aerobatic display includes a vertical dive, high drag will reduce the dive speed the aircraft can reach, so the inertial loads at the end of the dive will be lower. Also, lower speeds means that maneuvers can be flown more tightly for the benefit of spectators.
Other than that: No, less drag is always better.

This answer is seemingly saying that having lower lift to have less drag is better, even if it means having a lower lift to drag ratio. Can you clarify this confusion please?

@SharkyPanda I never touched lift in that answer. It only deals with drag. Lift is non-negotiable: It must at least be equal to weight; in turns even higher. Less drag with equal lift is better.

Well, by mentioning vertical maneuvers and specifically vertical dives, you kind of did touch lift-- would it be fair to say that the same characteristics that give high Drag in a vertical dive also tend to degrade  the max possible L/D ratio that can be attained in cruising flight, where Lift is high?  The point being that the tradeoff is acceptable in purpose-designed aerobatic biplanes etc--

@quietflyer: That is correct. Drag-producing designs tend to produce their drag with or without lift. In a pure vertical maneuver overall lift is zero, btw.

High lift at the expense of even higher drag means that the plane will not be able to fly very fast, as drag rises sharply with speed. But the extra lift is still useful in several situations and is often provided by drag-creating high-lift devices. Some of these situations include:

STOL (short takeoff and landing) and low-speed flight performance, where high lift at low speed is the critical capability. High-lift devices such as large slats and/or large flaps are deployed, then often retracted for cruise.

Fast climb rate, where surplus lift is essential but forwards speed less so. Some planes will partially deploy their flaps to improve climb rate.

Extreme high-altitude planes. All planes are limited in altitude because the stalling speed increases with altitude and eventually reaches the plane's maximum speed. Reducing the stalling speed, typically by adding wing area, allows the plane to fly higher. At lower altitudes these huge wings will be draggy, even though the lift is the same at all altitudes.

Re "Fast climb rate, where surplus lift is essential.... Some planes will partially deploy their flaps to improve climb rate.-- well, we're back to the old coefficient versus force thing, which the asker of the question has already expressed some puzzlement about.  A high lift coefficient may (?) help deliver a fast climb rate-- actually it is more important to a steep climb angle -- but lift is actually less than weight in a climb ( https://aviation.stackexchange.com/a/56476/34686). The suggested improvement being to consider replacing "lift" with "lift coefficient".

@quiet flyer I beg to differ here.  Aircraft that use wings for lift are far more efficient using them to climb than thrust.  Any climb angle you pick, more lift is lost from the wing than gained from the prop.  More thrust is used to maintain airspeed at higher wing AOA.  We need to get this squared away to continue improving the state if our art.  We can see raising AOA is what is actually done to climb.  The lowest drag combination of thrust and AOA produces the best climb.  One of your remembered quotes:  at cruise, I just pull a little stick to climb.  Why is this smart?

BTW we do have "curved flight" from excess lift in a turn, but we don't in a side slip (because of the cross wind). What prevents upward "curving" from excess lift in a climb?  Happy to continue working on this. We are getting better.  One needs excess vertical Lift to climb.  Upward acceleration is matched by vertical drag.  This is a mirror image of gliding is it not?

Let us continue this discussion in chat.

(I've added more comments in chat)

Whoever is upvoting "I beg to differ here"-- read the chat-- join in too if you wish--

OK, I'll do it your way.  Staticly stable. We won't touch trim, just add power.  The plane speeds up (excess thrust), then pitches up (more AOA!) until it settles at its trim speed in a climb. But no! The engine is not pulling "up", static stabilty still works here, it needs more Lift to hold straight line flight because the Lift vector is deflected from the vertical (just like shrinking your wing). So you need either more AOA or more speed to generate more lift. That's how it works.

A significant point about a high-lift wing in the climb is that the AoA is less than for a conventional wing. Either way, the engine needs excess thrust to overcome the extra drag. Which approach the designer takes is usually governed by other factors and other design decisions already made.

@Guy Inchbald that's the whole fuzzy thing about coefficient of lift factor, it combines AOA and wing type/configuration.  Generally, you see the bush planes with lots of horsepower to over come the drag of high lift configurations (gobs of camber, slots in front, slots in flaps), but it does get them off the ground at much lower V.  Really the antithesis of the graceful sailplane.

@RobertDiGiovanni -- not fuzzy; just the difference between a high lift coefficient, and a high ratio of lift coefficient to drag coefficient.  Many modern sailplanes actually have a rather low max lift coefficient, but a high ratio of cl/cd. As for the bush planes-- see last paragraph in my new answer to the present question.

The question shows some confusion around the difference between forces and their coefficients.
Let's address forces first.
The key thing about forces is that in an unaccelerated state (which excludes turning flight) we have to be able to rearrange the force vectors into a closed triangle, square, or other closed figure.
As in the vector diagrams shown in these ASE answers:
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?
Does lift equal weight in a climb?
Is excess lift or excess power needed for a climb?
You can see that in the simple case where the Thrust vector is parallel to the flight path (or is zero), Lift = Weight * (cosine glide or climb angle), so Lift is less than Weight if we're descending or climbing.  Lift only equals Weight when we are flying level.
We'll continue to keep things simple by assuming that Thrust vector acts parallel to the flight path, and therefore directly opposite to the Drag vector, throughout the remainder of this answer.  (See the last link above for more a more detailed treatment of the case where a significant upthrust or downthrust is in fact present, in relation to the direction of the flight path.)
For small climb or glide angles, the decrease in the Lift vector is small, but it is not zero.
Since a force is proportional to its coefficient * airspeed squared, there's no problem with the lift coefficient being higher in a slow climb than in fast cruise. Lift and drag coefficients are correlated to angle-of-attack. If we've increased the angle-of-attack, we know we've increased the lift coefficient.  But not the Lift force.  Excess Thrust, i.e. more Thrust than Drag, not excess Lift, is the key hallmark of a climb.
An excellent place to start learning more about lift and drag coefficients is section 4.5 of John Denker's "See How It Flies" on-line book.
See especially the green line on fig 4.14, lift coefficient versus alpha, fig 4.16, lift coefficient versus airspeed, and figure 4.17, Lift force versus airspeed.  You can see how in level flight, Lift stays equal to Weight, even though the lift coefficient is changing as angle-of-attack and airspeed change.
After finishing reading this answer, the reader may have a good idea of how to modify these graphs for climbing flight at some given climb angle.  The key changes are: the total Lift force is decreased by a factor equal to the cosine of the climb angle, and the airspeed for any given angle-of-attack is reduced by a factor equal to the square root of the cosine of the climb angle.  All because some of the Weight is borne by the Thrust vector rather than the Lift vector.
Many people find the idea that Lift is less than Weight in a climb to be very counter-intuitive.  What happens when we transition from high-speed cruise into a climb by pulling back on the control stick or yoke to slow to closer to Vx, with no change in Thrust?
The key hallmark of a climb is excess Thrust compared to Drag. If Thrust is constant, we have to reduce Drag if we want to climb1.  How can we do that?  By improving the L/D ratio.  We increase the angle-of-attack, so that the lift coefficient increases.  The drag coefficient increases too, but not as much, so the ratio of lift coefficient to drag coefficient improves.  The L/D ratio is arithmetically equal to the ratio of lift coefficient to drag coefficient, so the L/D ratio improves too.  When the airspeed is finished adjusting itself (decreasing) so that the Lift vector doesn't exceed the Weight vector, and more specifically becomes equal to Lift * cosine (climb angle), we find that Drag is now less than Thrust, and up we go.2
We can see from the vector diagrams in Does lift equal weight in a climb? that our climb angle will be equal to the arctangent of ((Thrust-Drag) / Lift)), which is also equal to the arcsine of ((Thrust-Drag) / Weight)).  Note the latter expression-- it is important to understand that the climb angle can be expressed in a formula that makes no reference to the Lift vector at all.  Optimizing the climb angle is all about maximizing the value of (Thrust-Drag).
A key point is that increasing the angle-of-attack doesn't actually improve the drag coefficient.  So how can it reduce the Drag force?  Because the increased lift coefficient causes a decrease in airspeed that causes a net reduction in the Drag force.
Of course, there's short interval immediately after we've moved the stick or yoke aft, but before the airspeed has had time to decrease substantially, where Lift is actually greater than Weight.  During this time, the flight path is curving upward into the climb.  This is an accelerated condition.  The curve may be so gentle that the pilot can't even feel the extra G-loading, but some extra G-loading is in fact present during the transition.  During this transition, we've also increased Drag.  The increased Drag force, plus the fact that as the flight path starts to curve upward, the Weight vector starts to gain a component acting against the direction of the airspeed vector, are responsible for the loss of airspeed during this transition into the climb.
With that background under our belt, let's turn our attention to some specific aspects in the original question.

From my understanding, high lift to drag ratio could make an aircraft efficient during cruising conditions. This is when the aircraft is in
equilibrium, lift is equal to weight, and thrust is equal to drag, and
since there's less drag, less thrust is required. Is this correct

Yes!

For example, if lift generated were to be around 10000N with 700N of
drag (14.29 l/d) compared to less lift and drag, 8000N of lift with
600N of drag (13.33 l/d), would less drag at the expense of less lift
be worth it in the context of cruising conditions (where thrust would
need to equal drag to maintain constant velocity)?

Here's the tongue-in-cheek answer:
Absolutely!  You've reduced Weight by 2000 Newtons, so you'll need less Thrust.  Your Drag force is now only 600 Newtons, so that's how much Thrust you'll need.  Reducing Weight is always helpful, if the goal is to minimize Thrust required in cruising flight.
But if the intention was that Weight actually remained constant, well-- hopefully by now the reader will understand that there's a flawed paradigm at play here.  If Weight is constant, then no matter what we do to the L/D ratio, we aren't free to vary the size of the Lift vector in cruising flight.  Lift is constrained to equal Weight.  But we can optimize the ratio of lift coefficient to drag coefficient, i.e. the ratio of Lift to Drag, so that Drag is minimized and the Thrust requirement is therefore also minimized.
On the other hand, as other answers have noted, a low ratio of lift coefficient to drag coefficient, and therefore a low ratio of Lift to Drag, is beneficial to make the glide path steeper during approach.  But we also want a low airspeed, and therefore a high lift coefficient, as we approach to landing.  These goals are not inconsistent-- flaps will deliver them nicely.  Flaps increase the lift coefficient, but they increase the drag coefficient even more.
In hindsight the idea of reducing Lift (without reducing Weight) in cruising flight may seem a bit foolish.  But the truth is that pilots very often envision that Lift is greater than Weight in a climb, which is an equally flawed concept.3  So the confusion is understandable.  What we actually often increase to enter a climb is not Lift, but lift coefficient.  And the fundamental reason that we do this is not because we need more Lift, but rather because we want a better ratio of lift coefficient to drag coefficient and therefore a better ratio of Lift to Drag.  All because we need to reduce Drag, to maximize our climb angle for a given amount of Thrust.
Likewise, pilots often speak of a "high-lift wing".  It would be more clear, and might help to avoid some of the confusion contained in the original question, to speak of a wing with a high maximum lift coefficient.
A related question-- what is going on with a STOL bush plane with slats, lots of wing camber, etc-- if climbing is all about minimizing Drag rather than maximizing Lift, then how can all that Drag help the climb angle?
Answer-- the main purpose of those design features is to increase the lift coefficient and help the aircraft land and take off slowly.  If thrust is constant, then the aircraft's maximum climb angle occurs at the maximum ratio of Cl/Cd and L/D, where Drag is minimized and (Thrust-Drag) is maximized.  Many of those design features increase the drag coefficient so much that they decrease the maximum ratio of Cl/Cd and L/D.  However, in the real world, for piston and turboprop engines, substantially more thrust is available at lower airspeed than at higher airspeed, so some of those features may offer a net improvement in the maximum climb angle after all.4  Consider also that there's a benefit in obstacle clearance if the maximum climb angle is achieved soon after takeoff, rather than after a long period of acceleration.  But the fundamental reason that those features are there is to permit slow--i.e. short-- takeoffs and landings.
Footnotes--

Of course in reality, Thrust does not actually stay constant as we increase the angle-of-attack and slow to a lower airspeed, especially in a piston-engine aircraft, which tends to produces a roughly constant amount of power.  More Thrust typically becomes available as we reduce the airspeed, which is another reason why the climb angle improves as we slow down to get closer to the Vx airspeed.

See footnote 3 for a completely different-- and erroneous-- description of what happens when we move the stick or yoke aft to start climbing, taken from FAA flight training materials.

For example, on page 3-16 of the FAA's "Airplane  Flying Handbook" (2016), we read "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb. An airplane can only sustain a climb when there is sufficient thrust to offset increased drag..."  As discussed in this answer, this is simply not true.  If we transition from level flight to a climb without doing something (like slowing to Vy or Vx) to reduce the Drag force to a lower value than we had a level flight, we certainly will have to increase Thrust, but that that excess Thrust is being used to help support the aircraft's Weight, not to counteract increased Drag due to increased Lift.  FAA ground school training materials of this nature are often found to be poor sources of factual information about the actual forces present in various in-flight scenarios.  See for example the poor depiction of the forces in gliding flight in the "Glider Flying Handbook" (2013) as discussed near the end of this related ASE answer, and the poor depiction of the forces in slipping or skidding flight in the "Pilot's Handbook of Aeronautical Knowledge" (2016) featured in this related ASE question.

Sometimes so much power is available that the maximum available climb angle is simply not an issue of concern-- watch bush plane on steroids "Draco" on this You Tube video-- but still note the difference in configuration between approach and landing and takeoff and climbout.)

Much of this answer could be well adapted to a potential ASE question "What exactly happens to the magnitude of the lift vector as we transition from high-speed cruise into a moderate climb by moving the stick or yoke aft slightly, while holding Thrust exactly constant"? Staying faster than max L/D speed throughout. Maybe someone should ask that question. Maybe someone would like to attempt to offer an alternative theory of what happens in such a case.

Such a question might arguably be a duplicate of http://aviation.stackexchange.com/q/40921/34686, except that question specifically says it is really asking about the sum of all vertical forces , which is not actually the same as Lift.

If the OP insists on the absolute numbers (as opposed to coefficients), then the paragraph in bold is the answer, and there is nothing cheeky about it. I was about to post it until I saw yours, but I fear not everyone will reach that part... ;)

@Zeus -- "but I fear not everyone will reach that part... ;"  -- truer words have never been spoken--

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text:
What is physics explanation for minimum sink rate airspeed?
I did some research on Minimum Sink Rate, and saw much information defining it, (such as it is the speed at which the aircraft will remain in the air for the longest time, etc.) and how it is generally lower than Best Glide speed, and is generally a few knots above stall speed. There was also some information about what is used for (when you are trying to remain airborne for as long as possible after engine failure in order to deal with an emergency, or for gliders to maximize the climb effect from a thermal),
…. But I could not find, anywhere, an analysis/explanation for how this works, from a Physics perspective, as you can easily find to explain Vx, (best angle of Climb airspeed), and Vy (best rate of climb airspeed), that explain/analyze them from a Physics perspective (Vx is speed at which you have the highest Excess Thrust, whereas Vy is the speed at which you have Maximum Excess Power).
Also, from my limited understanding, what I did read seems to violate my basic understanding of what Min Sink Rate means, from a Physics Perspective. As I would understand it, Min Sink Rate Speed is the speed at which your rate of descent is the lowest, (you will remain airborne for the greatest amount of time). This is equivalent to the airspeed at which the aircraft is losing altitude, (Potential Energy), at the slowest possible rate. With no power on the aircraft, loss of energy (altitude) is directly related to total drag, (energy must be conserved!), and this occurs at L/Dmax.  All the complexities involved with determining Maximum Excess Power, or Maximum Excess Thrust, as are required to analyze Vx and Vy, become moot.  So although it appears to be accepted wisdom that Minimum sink is different from Best Glide, this seems to violate basic physics. They should both occur at the airspeed (AOA actually) where total Drag is minimized, i.e., at L/Dmax. Where am I going wrong?

From first principle, assuming the airplane is a point mass, no wind, small angle of attack, and thrust acts in-line with drag:
$$T-D-W\sin \gamma=m\dot{V}$$
where $T$ is thrust, $D$ is drag, $\gamma$ is angle of climb, $m$ is mass of the airplane, $W=mg$ is weight, $V$ is airspeed/forward speed (assuming no wind).
In a total power off scenario, $T=0$, and assuming steady-state, $\dot{V}=0$:
$$\sin \gamma = -\frac{D}{W}$$
The rate of climb ($\dot{z}$), which is the negative of sink rate, is related to angle of climb by: $\dot{z}=V\sin \gamma$. Therefore, we have:
$$\dot{z}=-\frac{DV}{W}=-\frac{P_R}{W}$$
Therefore, for minimum rate of descent, we would like the minimum power required ($P_R$).
Note: for minimum glide slope, it corresponds to $L/D_{max}$, but not for minimum rate of descent.

So power required is product of total drag times velocity. And min rate of descent occurs when this product is at minimum. So to see where this occurs you need to express D as function of V and then determine value of V where this product expression is at minimum.

i.e., set dP/dV = 0 and solve for V.

You are right that in order to achieve minimum sink rate, you need to minimize energy losses. And yes, energy losses are related to the drag. But here you need to stop and think again what 'related' exactly means.
The drag is a force, not an energy. But when you multiply the force acting on moving body by its speed, you get power (that is energy per unit of time) inflicted by this force. (Actually, you have to multiply only the fraction of force, which is parallel with the velocity vector, but drag is, by definition, acting directly against movement, so this is fulfilled automatically).
Therefore energy losses are drag multiplied by speed. At L/Dmax you are flying with  minimal drag force acting on the airplane (for steady flight), but energy "consumption" is not at the minimum.
If you decrease airspeed a bit, drag force increases by some small amount, but the product of airspeed and drag decreases thanks to decreased speed. So you keep slowing down until these two effects cancels out and you end up at the point with minimal energy loss rate. That is the least "Watts" of drag, not "Newtons".

Thanks. This is exactly in line with JZYLs answer. I gave him the check because he included mathematical derivation.

@CharlesBretana Of course, accept the answer which works best for you. I have seen JZYLs answer in middle of my writing, but wanted to post my one too, so there can be another explanation in different words.

In addition to some excellent content in other answers, it's worth noting that V-bestglide occurs at the angle-of-attack where L/D (and therefore also Cl / Cd) is maximized, while for shallow glide angles, it's a good approximation to say that  V-minsink occurs at the angle-of-attack where (Cl^3 / Cd^2) is maximized1.  The difference between the two formulae means that V-minsink will always occur at a lower airspeed than V-bestglide.  In fact the reduction in sink rate obtained by slowing down is the reason that the Cl term is cubed, while the Cd term is only squared, in the expression for V-bestglide.  This is related to the concept that required power is minimized at V-minsink.

loss of energy (altitude) is directly related to total drag

No, we can show that total Drag is minimized at V-bestglide, but we if fly more slowly down a slightly steeper glide path, we can end up with a lower vertical speed.  That's the reason for the difference between the airspeeds for min sink and best glide.
Footnotes--

Derivation: the glide ratio can be geometrically be shown to be equal to L/D, which is arithmetically equal to Cl/Cd.  The sink rate is proportional to airspeed * (1/ glide ratio), which is the same as airspeed * D/L or airspeed * Cd/Cl.  At shallow glide angles, nearly all the Weight is supported by the Lift vector, so it's a good approximation to say that Lift is constant and airspeed is therefore inversely proportional to the square root of Cl.  Therefore at shallow glide angles, it's a good approximation to say that the sink rate is proportional to Cd / CL * 1/(Cl^1.5), which works out to Cd / (Cl^1.5), which is the same as (Cd^2 / Cl^3).  A derivation of this appears in "Model Aircraft Aerodynamics" by Martin Simons (3rd edition, 1994) on pp. 40-41, or in more detail, on pp. 238-239. (link to PDF).

Related content appears in this ASE answer--
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?

This answer would benefit from adding some content that is not limited to the case of shallow glide angles. For example, when the glide angle is very steep, the sink rate is proportional to the square root of Cd (think of a parachute). There is a formula that gives the sink rate at all Cl/Cd ratios, but it has not yet been included in this answer. Stay tuned


text:
What is the decrease in power when a drone is in descent?
For a quad-copter, or a fixed-wing drone with quad-rotor formation, how much would the power of the motor decrease when a constant speed descent is needed to bring the drone back to the ground? Does the motor just need to ensure Lift equals the Weight so the drone can descend in constant speed?
Or, in other words, can the the motor power during descent be calculated using the climb power requirement equation, but just putting in negative speed to represent descent?

Off-topic here. Ask in the Drones and Model Aircraft StackExchange.

On-topic here because the FAA considers all drones to be "Unmanned Aircraft Systems" --  see https://www.faa.gov/sites/faa.gov/files/2021-08/RemoteID_Final_Rule.pdf

We still have an issue here about whether you really intend to ask whether Lift equals Weight minus the vertical component of Drag-- I don't want presume to edit the question, but you may wish to --  but this really  this pertains to thrust rather than power --

@quietflyer, a good case could be made for either, but your quote equally bolsters an opposing view.  Consider:  BECAUSE the FAA considers all Unmanned Aircraft Systems to be drones, and BECAUSE Stack Exchange has a separate site for drones, THEREFORE the Drone SE site is the "best" place for this question...

There may be multiple questions here --answer may be very different for quadcopter w/ no fixed wing vs quadcopter that also has a fixed wing.

can motor power during a descent be calculated from the climb power requirement?

Usually it's the other way around, but most importantly, one is quantifying rate of climb.
Power requirement for level flight is determined in fixed wing craft by the glide ratio.  One you know your altitude loss over a certain distance, power (Thrust at glide speed) requirement for level flight is mass × glide ratio.
Climbing ability is determined by excess thrust available at that speed.  As one angles up, lift requirement (perpendicular to the line of flight) becomes less than weight (by the cosine of climb angle × weight) and thrust must not only maintain airspeed but also contribute to vertical force to support the weight against gravity.
For a vertically descending helicopter, quad, landing Starship, whatever, the physics is a bit easier; it is just based on thrust, drag, and gravity.
What one must consider here is rate of descent.  Aside from "lawn darts", most vertically descending objects will reach a "terminal" constant rate descent velocity from drag alone.  Adding vertical thrust force will slow the rate of descent.
In a hover, vertical thrust force = weight.  Ascending, vertical thrust = weight + drag (dependent on rate of ascent).  Descending, vertical thrust = weight - drag (dependent on rate of descent).
Actual power requirement may also depend on technique of descent.  Your quad may require less power (for a given rate of descent) if it is moving forward, as opposed to straight down, just like a helicopter.  The forward motion actually helps it "glide" a bit.

In conventional fixed-wing airplane, lift decreases (slightly) during climb or descent, and climb is due to excess thrust, descent due to lack of thrust.  But in quadcopter, how do we define thrust?  Wouldn't it be vertical in a vertical descent?  And wouldn't it almost always point in quite a different direction than the direction of the flight path?  I'm not sure that answers from fixed-wing airplanes readily translate over to this situation.  Also answer may be very different for quadcopter w/ no fixed wing vs quadcopter that also has a fixed wing.

@quietflyer well, one can reason a quad is simply 4 rotating wings that do not require forward motion to generate what we define as lift.  "Quadracopter w/no fixed wing"  still benefits from forward motion.

For a helicopter the physics is actually much the same as for airplane. Note that auto-rotating helicopter – which is pretty much equivalent to fixed-wing gliding, it's just several wings gliding in circles – requires no power, but most of it's lift is still provided by the rotor, not drag on the body.

Absolutely, and, in the end, it takes a certain amount of power to fly, and more to climb.

Lift will be less than weight in a descent. This results in a vertical speed component, that builds up until the vertical drag equals the force difference. Like any falling object reaching a terminal speed.

Or, in other words, can the the motor power during descent be calculated using the climb power requirement equation, but just putting in negative speed to represent descent?

Yes indeed, provided that vertical drag is a part of the climb power equation. Motor power is a funtion of generated thrust.

Motor power is function of generated thrust and the velocity in the direction of thrust! After all, during autorotation the rotor still produces thrust to balance weight with no power input at all.

The power required to maintain lift equal to weight¹ increases with climb rate and decreases with descent rate.
Thrust produced by a rotor (propeller) is equal to mass flow rate of air through the propeller disk times change in velocity of that air. This change in velocity increases momentum of the air (that balances the thrust), but it also increases its kinetic energy, and this energy has to be provided by the engine as induced power³.
But kinetic energy is equal to mass times square of velocity, which means the induced power is equal to mass flow rate times change in velocity times velocity² of the air. This velocity is some base value due to the rotor working, plus the vertical velocity of the craft.
So if the craft is climbing, the required power increases, and if the craft is descending, the power requirement decreases, even for the same lift.
If the angle of attack of the rotor blades is low enough so they don't stall when the flow direction changes to upward, the induced power will decrease below zero until it balances the parasite power³ and the rotor spins without any engine power at all. This is called autorotation, and allows helicopters to still make a safe landing in case of engine failure. However it requires the blades to have low enough angle of attack, so a quadcopter with fixed blade pitch won't autorotate, but the rotor will stall at higher rates of descent instead.

¹ Drag will increase the lift requirement in climb a bit, and decrease it in descent a bit, but with typical climb or descent rates – 100 ft/min ~ 1 knot – this is relatively small contribution.
² More precisely, $P = \dot m (v_0 \Delta v + \Delta v^2)$ where $P$ is the induced power, $\dot m$ is mass flow rate, $v_0$ is the air flow speed just above the rotor and $\Delta v$ is the increase in flow speed through the rotor. When the craft is descending, at some point $v_0$ will become negative, meaning the air flows from below.
³ Power required to turn a rotor (propeller) is induced power, which is the power required to increase kinetic energy of the air to produce thrust, plus parasite power, which is the power required to overcome parasite drag of the blades.

Any decrease in power below what is required to maintain level flight will result in a descent.


text:
What is the effect of thrust vectoring effect on the rate of turn?
I just got a question after reading my old Principles of Flight book. The equation for the rate of turn is $\textrm{ROT} = 1,091 \cdot \frac{\tan(\phi)}{V}$,
with $\phi$ denoting bank angle and $V$ denoting airspeed
How would that change when the airplane is equipped with thrust vectoring like the F22 or SU35? Is there an equation for that?

If you want to get down to actual equations, you are going to specify HOW the thrust vectoring is being used.  Say we are in a left turn.  Is the nozzle pointed to the left, to drive the nose to the left and expose the right side of the fuselage to the airflow?  Or is the nozzle pointed to the right, to provide centripetal force to increase the turn rate?  If the latter, what (if anything) is providing the yaw torque needed to keep the nose from yawing to the right?  The question as you are asking seems to be insufficiently constrained.

Above was for the Su-35 , w/ 3-D thrust vectoring.  F-22 only has 2-D.

An interesting case that may provide some food for thought in improving the question is Northrop YA-9.  https://en.wikipedia.org/wiki/Northrop_YA-9 . Airbrakes on wings could be asymmetrically deployed to facilitate using the rudder to generate aerodynamic sideforce. As an example, to the boost the turn rate in left turn, without increasing bank angle, the plane might deflect rudder to right to generate an aerodynamic sideforce to the left. Meanwhile the nose would be prevented from yawing to right, and kept aligned with the instantaneous flight path, by opening the airbrake on left wingtip.

So are you envisioning something like this that combines some sort of differential wingtip drag device, with the thrust vectoring nozzle? (Does the Su-35 actually do this? Not as far as I'm aware, but maybe-- perhaps there's a good basis for another ASE question here--)  As an aside, my understanding is that the basic purpose of the system on the YA-9 was to allow gun-aiming and flight path control to be de-coupled.

The standard equation applies to any coordinated level turn (so that the G felt is "straight down" to an occupant of the aircraft, in an aircraft frame of reference). Matters not how it's all achieved - flaps, high-lift devices, helo rotor, thrust vectoring, whatever... a coordinated level turn at "this" bank will take "this" much radius. And it will have a calculable G load.
Actually achieving (and sustaining) that G load may take all sorts of interesting effects, such as thrust vectoring. But once you're there, the standard equations apply.
And, if you choose to fly an un-coordinated turn, with some amount of side-loaded G force (example, a car turning at high speed... no appreciable "bank angle" so the G force pushes you toward the outside of the turn), then all the standard assumptions are out the window. At that point, your radius would be a function of G & not bank... and it'd get uncomfortable really fast!

I know we aren't supposed to use comments to say "nice answer"-- but this is a nice answer.  Nicely sidesteps (yet also illuminates) any issues due to question being insufficiently constrained.

Not to be pedantic but in circular motion there is no "force pushing you towards the outside of the turn" you are only accelerating inwards.

@bruh_weed: True, the actual force is pushing the plane inwards, not pushing you outwards. But it feels the same either way — you and your seat are still being pushed together.

With thrust vectoring you no longer turn (as in: the wing creates the force that accelerates you in the desired direction) but you do post-stall maneuvering. Next, you need to distinguish between highest instantaneous turn rate (trading altitude for higher rate) and continuous turn rate (which is limited by the available thrust in most cases).

Turn rate diagram (picture source). It plots flight Mach number on the X axis over turn rate on the Y axis. The bold colored lines show the sustained turn performance of several airplanes. At low speed the turn rate grows in proportion to the maximum possible load factor that the maximum lift of the airplane is capable of. The kink in the lines at around 10 - 12 degree per second shows the thrust limit - in order to fly tighter turns at even higher load factors, more than the installed thrust is needed. Now the curves run almost horizontal along the thrust limit and decline again at high Mach numbers, first at the speed of sound and then when supersonic drag cuts down the possible load factor.
The thin colored lines show the instantaneous turn rate, when altitude loss is permitted. The sharp peak at the maximum load factor (for example 8g for the Su-27 and 9g for the F-15) marks the maximum turn rate when thrust is unvectored and wing lift is used to force the change of direction.
With thrust vectoring the airplane can fly a direction change in a totally different way. It will pull up to reduce speed, then rotate using vectored thrust when flying at low speed in a zero-g parabola. When the fuselage points at the desired direction, it will use the altitude gained in the pull-up to accelerate again, now in the new direction. Now the turn rate depends on how quickly the speed can be reduced and how long it takes to build up new speed. The rotation itself only needs a second or two.
Since a conventional turn also requires to slow down and missile engagement can start as soon as the fuselage points towards the adversary, post-stall turns using thrust vectoring give a decided advantage in a dogfight with missiles.


text:
How did the Wright plane takeoff/land?
I don't understand exactly how the Wright brothers handled takeoff with the first gliders and powered plane. I know it has something to do with an elevated rail track. Also, if the flying position was head first, body parallel to the ground, how did they land without injury?
Separate question: For the initial glider flights, did they start at a high elevation and simply glide to the ground, or was there any method for increasing altitude during flight?

Re : "I don't understand exactly how the Wright brothers handled takeoff with the first gliders and powered plane. I know it has something to do with an elevated rail track. ". -- please consider editing to transform these two sentences into an actual, answerable question--.  (PS please note that the monorail track was not used for any of the gliders that preceded the 1903 powered airplane.).

(Sorry, intended answer got mis-posted as edit, now rolled back)

As you said, they used a monorail for launches. The aircraft was pulled forward by a pulley which was connected to a weight which hung from a derrick. First the weight had to be pulled up, the pulley was run to the tip of the rail and back to the Flyer, and once the engine was running and the pilot in place, a helper released the weight and the pulley pulled the aircraft forward.

The pictures here are taken from this site; I recommend it for further reading!
Below is a picture of the rail, note the uptick at the end which gave the airplane a gentle push upwards, similar to the ski jump technique of British aircraft carriers.

For landing the Flyer used two skids. Depending on the ground, this could mean a gentle slide to standstill. Rougher landings would mean bruises for the pilot, like the ones suffered by Orville on August 23, 1904. But in most cases of a rough landing, the elevator assembly absorbed most of the impact energy. Elevator repairs were quite frequent.
I know less about the glider launching technique. Most glider tests were done with an unmanned glider held by strings into the wind. Glider flights would be downhill; thermalling was discovered much later (in 1913). However, even Otto Lilienthal already mentioned the uplifting effect of gusts, but neither he nor the Wright Brothers could circle tightly enough to stay inside a thermal.

For their first flight 17 Dec 1903 they used a rail for takeoff, but NO CATAPULT. The Kill Devil Hills, North Carolina site was chosen for strong steady winds, so they DIDN'T NEED a catapult. And their first flight patent claim was to take off without external assistance, demonstrate stable controlled flight, and land at a site no lower than the takeoff point.
The catapult was used only after their experiments were moved to Huffman Prairie near their home in Dayton, because the winds were less than in North Carolina.

Thanks for the detail! As the caption on the picture of the derrick says, a catapult was only added from September 1904 on. However, the majority of flights (also all flights in Europe) were with the catapult.

Separate question: For the initial glider flights, did they start at a high elevation and simply glide to the ground, or was there any method for increasing altitude during flight?

The Wrights did experience soaring flight, including altitude gains to higher than the take-off point, while flying in conditions that we now would call "ridge lift" or "slope lift", on the sand dunes of Kitty Hawk in some of their gliders before their 1903 powered flight.  In such cases the altitude gains were due to the prevailing meteorological conditions, not the method of launching.

and... this is why there should only be one question per question. Great answers to the "other" question can't be accepted. :/


text:
Reasoning Behind Individual Airspace Designations
Is there a site that explains the reasoning behind each individual airspace designation?
I ask this because the sectional chart shows a 6 NM radius area centered at about 40°22'30.0"N 77°37'30.0"W where class E airspace begins at 700 feet.  Neither the sectional chart, nor Google maps, nor Bing maps show anything of interest there; and nobody I asked seems to have a clue, either.
Here is the area in question:

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Is there a site that explains the reasoning behind each individual
airspace designation?

No-- unless you consider the "Federal Register" to be a "site".   The reasoning behind any given individual airspace designation will be explained in NPRMs (Notices of Proposed Rule-Making) and Final Rules set forth in the Federal Register.  However, it takes some effort to find this content.
I'll walk the reader through the steps I took to learn the reasoning behind the specific airspace designation in question.

Got to the following link: https://www.faa.gov/uas/commercial_operators/uas_facility_maps.  This is a link that is intended for use by "drone"/ sUAS/ model airplane operators, but is also highly useful for our current purpose.

Under the "Quick Links" header, click on "All UAS Facility Maps And grids".  (The link you are clicking on is often called the LAANC map or LAANC grid.  LAANC stands for "Low Altitude Authorization and Notification Capability".)

This (or a similar) URL should come up: https://faa.maps.arcgis.com/apps/webappviewer/index.html?id=9c2e4406710048e19806ebf6a06754ad

On the upper right side of the task bar, click on the "Layer List" icon.  Click "Class E5 airspace" on, and click all other options off.  (Most, if not all, Class E airspace that terminates with a floor at 700' AGL is classed as "E5" airspace.)

Browse around the map till you find the particular airspace structure of interest.

Click on that airspace structure.  You should get an inset page that has some data about the airspace.  In this particular case, the page includes a line that reads "NAME:  HONEY GROVE CLASS E5"

Armed with that information, we need to search the latest version of the FAA's "Airspace Designations and Reporting Points" document (direct link to PDF:  https://www.faa.gov/documentLibrary/media/Order/2022-08-19_FAA_Order_JO_7400.11G.pdf) to see when that airspace was established.  Scrolling through the "E5" airspace section, we find the relevant entry on page E-384.

The document states that the airspace established is "That airspace extending upward from 700 feet above the surface within a 6-mile radius of the point in
space for the SIAP to the EWT 4 Heliport, Honey Grove, PA."-- which tells us why the airspace is not centered around the heliport itself.  This appears to be the same heliport that is identified as "Stottle Memorial" on current sectional chats.

The document also states "AMENDMENTS 10/03/02 67 FR 19108", telling us that the airspace was last amended 10/03/02, on page 19108 of the 67th volume of the Federal Register.

A logical next step now is to type "67 FR 19108" into our browser window.

In the links that come up, the second-to-top one is entitled "67 FR 19107 - Amendment to Class E Airspace; Caruthersville".  This is only one page off from our target.  The next step is to click on "document in context", and then select "view entire issue PDF", and then scroll down to our target page, 19108.

(Here is a shortcut to that PDF:  https://www.govinfo.gov/content/pkg/FR-2002-04-18/pdf/FR-2002-04-18.pdf .  Scroll down to page 19108.)

The title of this section is "Establishment of Class E Airspace;
EWT 4 Heliport, Honey Grove, PA".  The specific rationale for establishing this Class-E-to-surface airspace is described in more detail in the text that follows.

The essence of the matter is encapsulated in the following text: "This amendment to Part 71 of the Federal Aviation Regulations (14 CFR Part 71) provides controlled Class E airspace extending upward from 700 feet above the surface for aircraft conducting Instrument Flight Rules (IFR) operations at the EWT 4 Heliport, Honey Grove, PA."  Also:  "This action establishes Class
E airspace at EWT 4 Heliport, Honey
Grove, PA. Development of Standard
Instrument Approach Procedure (SIAP)
based on the Global Positioning System
(GPS), Helicopter Point in Space
Approach at the EWT 4 Heliport, has
made this action necessary. Controlled
airspace extending upward from 700
feet Above Ground Level (AGL) is
needed to contain aircraft executing the
approach to the EWT 4 Heliport." Further down in the "Final Rule" we find the exact same text that appears in the "Airspace Designations and Reporting Points" document (see point #8 above).  Perusing the NPRMs (Notices of Proposed Rule-Making) that preceded this Final Rule, also published in the Federal Register, might provide additional context.  Federal Register citations for the relevant NPRMs are embedded within the text of the Final Rule.

You can also search the Federal Register directly, once you've obtained the name of the E5 airspace from the LAANC site, but this is probably a better way to go--

Fantastic answer... teach a person to fish and give them the fish!

@quiet flyer:  Thanks! Exactly what I was after.  Maybe the location of the PinS ensures helicopters flying under IFR do not strike the nearby power lines.  As I understand it, a helicopter following IFR flies to the PinS and then proceeds to the landing location under VFR.  Requiring VFR to the landing location would minimize the chance of hitting the power lines, since the pilot is forced to cross over them on the way to the landing spot.  I am a student sport pilot and welcome corrections to my understanding.  I know everything I just said may be either stupid or common knowledge.

Some useful info on the rational of helicopter Point In Space approaches has been added in chat, with useful links (but not specific to this particular heliport) -- see this chat post and following ones-- https://chat.stackexchange.com/transcript/message/62915064#62915064


text:
Do minors need parental permission to fly with a private pilot?
I’m a 17 year old private pilot, and my friend and his girlfriend who are also 17 want to fly with me. It’s not clear whether they are asking their parents permission. If their parents didn’t want them to go and find out, would I, the pilot, get in trouble? Are there any regulations regarding this subject? Should I get their parents written consent before taking them flying?  This question pertains specifically to the United States.

The answer to your question entirely depends on where in the world you are.

It would be in the US

Please note that we can only answer about aviation regulations (FAA in this case) on this site. There might be other laws and regulations about transporting minors. Questions about such local laws should be asked on Law.SE.

It probably also (or only) depends on the law of your state.

It very likely (for state law) also matters whether you'll be landing back where you took off, or somewhere else in the same or another state -- and in the latter case, might depend on what the minor couple get up to after landing (or during the flight).

Someone needs to edit to say "In the US"  ---

@quietflyer I added the [faa-regulations] tag based on OP's comment above. IMHO there is no need for spelling it out in the question. If you think differently, feel free to edit it in.

@Bianfable -- since anyone can edit tags, I think it's best that the question be self-sufficient independent of whatever tags are attached-- editing--

In the state of California, it's technically considered kidnapping to transport a minor across county lines against their parents' permission.

@Phil Miller, "against parent's wishes", or "without parent's permission"?  Those are two very different legal standards.  And FWIW, if all parties are minors it is likely different than if one is over 18.  Things get messy when a 17 year old "child" and their best friend who is a week older at 18 (and therefore an "adult") get into a legal scrape...

I assume you are not the estranged partner of the other parent of one of the minors, and your route is not crossing national borders ;-).

There’s no FAA regulation requiring that. That being said, if your passengers are minors, their parents have the ultimate say as to what they can and cannot do.  It would be wise to make sure that your friends have permission from their parents for the activity prior to taking them as passengers.

Sorry for taking this off tangent, but I just had to express my bemusement by your implication that parents have any real say about what their 17 year old kids can and cannot do ;)

@yoniLavi Legally they do, and the person transporting them without permission (by any means) can be in serious legal trouble.  But this is a (complicated) legal question and that is why the question would be better asked at https://law.stackechange.com.

Well they do.  And unless there is a compelling state interest otherwise, the authorities will side with the parents.

@RBarryYoung really? So a 17 year old who boards a bus when forbidden to do so by his parents could get the bus driver into trouble? On the metro? An uber driver? So how is a private pilot different?

@abligh Seems like the difference would be that a bus driver has no knowledge of that, and no expectation to need to know, whereas the private pilot very obviously does know since they're asking about it here :p

@abligh The difference is not the means of transportation but the circumstances of travel, the relationship between the parties, and the relative ordinariness or unusualness of the circumstances. It's also almost definitely going to be a requirement of either whoever you are renting the plane from and/or your insurance.

Remember, the pilot is 17 also!  High school friends ride together in cars all the time, what is different about an airplane?

If the parents explicitly forbid the activity and you go against their wishes and take their child flying, you can be in big trouble for that

@abligh as I said "can be", yes, depending on the jurisdiction, the circumstances and their knowledge of the situation.  It's complicated.

If you are asking about FAA regulations, the answer is "no, there are no regulations on this subject".  I've been involved in EAA Young Eagle Rides, CAP glider encampments, etc, and if there had been any FAA regulations bearing on a requirement for obtaining parental permission before giving young people rides in aircraft, someone would have mentioned it at some point.  Not to suggest that parental permission wasn't obtained anyway, for other reasons-- but a specific FAA regulation on the subject would surely have been mentioned if it existed.

You don't need their parents explicit consent anymore then you would to give them a lift in your car, but you do need to ensure you are legally covered. As PiC you have legal responsibilities and liabilities so before you can take on the risks associated with carrying passengers you should check with whomever would be sued if they got hurt. If you are legally a minor, that would be your parents and/or whoever pays your insurance premiums.
Also be a little careful about crossing state lines, as the laws relating to transporting minors over state lines vary from state to state and can be quite surprising.